Solve the problem.
-In one U.S. city, the quadratic function $f ( x ) = 0.0037 x ^ { 2 } - 0.48 x + 36.78$ models the median, or average, age, $y$, at which men were first married x years after 1900 . In which year was this average age at a minimum? (Round to the nearest year.) What was the average age at first marriage for that year? (Round to the nearest tenth.)
A) $1965,21.2$ years old
B) $1965,52.3$ years old
C) $1936,52.3$ years old
D) 1954,36 years old

Question

Quizplus · Accepted Answer

The minimum value of a quadratic function $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$. For the given function $f(x) = 0.0037x^2 - 0.48x + 36.78$, $a = 0.0037$ and $b = -0.48$. Plugging these values into the formula gives $x = -\frac{-0.48}{2(0.0037)} = \frac{0.48}{0.0074} = 64.86$, which rounds to 65 years after 1900, or 1965. To find the minimum average age, substitute $x = 65$ into the original equation: $f(65) = 0.0037(65)^2 - 0.48(65) + 36.78 \approx 21.2$.

Solve the Problem f(x)=0.0037x2−0.48x+36.78f ( x ) = 0.0037 x ^ { 2 } - 0.48 x + 36.78f(x)=0.0037x2−0.48x+36.78

Solve the Problem $f ( x ) = 0.0037 x ^ { 2 } - 0.48 x + 36.78$